Question

Motion along a Circle A wheel of radius 2 ft makes 8 revolutions about its center every second. (a) Explain how the parametric equations \(x=2 \cos \theta, \quad y=2 \sin \theta\) \(x=2 \cos \theta, \quad y=2 \sin \theta\) (b) Express \(\theta\) as a function of time \(t\) . (c) Find the rate of horizontal movement and the rate of vertical movement of a point on the edge of the wheel when it is at the position given by \(\theta=\pi / 4, \pi / 2,\) and \(\pi .\)

Short answer

The expression for \( \theta \) as a function of time is \( \theta(t) = 16\pi t \). The rate of horizontal and vertical movements at \( \theta = \pi/4, \pi/2, \pi \) can be found by substituting these values into the derived expressions \( -32\pi\sin\theta \) and \( 32\pi\cos\theta \) respectively.

Step-by-step solution

Express \( \theta \) as a function of time.

Given that the wheel makes 8 revolutions per second, one revolution corresponds to \( 2\pi \) radians, therefore the wheel covers \( 8\times2\pi \) radians per second. Hence the angle \( \theta \) covered as a function of time \( t \) is \( \theta(t) = 16\pi t \).

Find the rate of horizontal movement.

The rate of horizontal movement is given by the derivative of the x-coordinate with respect to time, \( \frac{dx}{dt} \). The x-coordinate is \( x=2\cos\theta \). Therefore, \( \frac{dx}{dt} = \frac{d}{dt} (2\cos\theta)= -2\sin\theta(\frac{d\theta}{dt}) = -32\pi\sin\theta \). Calculate \( \frac{dx}{dt} \) at \( \theta=\pi/4, \pi/2, \pi \).

Find the rate of vertical movement.

Similarly, the rate of vertical movement is given by the derivative of the y-coordinate with respect to time, \( \frac{dy}{dt} \). The y-coordinate is \( y = 2\sin\theta \), hence \( \frac{dy}{dt} = \frac{d}{dt} (2\sin\theta) = 2\cos\theta (\frac{d\theta}{dt}) = 32\pi\cos\theta \). Calculate \( \frac{dy}{dt} \) at \( \theta=\pi/4, \pi/2, \pi \).

Key concepts

Circular Motion in Calculus

Circular motion in calculus is a classic application of parametric equations, where the position of an object moving in a circle can be described using trigonometric functions. In this context, the circle is typically centered at the origin of a coordinate system, with a point on the circle at any given time being represented by coordinates \(x(t), y(t)\). For a circle with radius \(r\), the parametric equations generally take the form:

\[ x(t) = r\cos(\theta(t)) \
y(t) = r\sin(\theta(t)) \
\]
Here, \(\theta(t)\) represents the angle in radians that is subtended by the radius and the positive x-axis as a function of time. In the specific exercise, a wheel of radius 2 feet exhibits circular motion, and the parametric equations are provided as: \(x=2\cos(\theta)\) and \(y=2\sin(\theta)\), which correspond to a circle with radius 2. Exploring circular motion through calculus allows us to analyze not just the position but also the velocity and acceleration of an object at any point in its path.

Rate of Movement Differentiation

The rate of movement differentiation involves finding the velocity of a point in motion using calculus. Here, we differentiate the position function with respect to time to obtain the velocity. For parametric equations, this process becomes the differentiation of each coordinate function with respect to time.

Considering the wheel moving in a circle, the horizontal and vertical rates of movement can be found by taking derivatives of \(x\) and \(y\) with respect to time. The derivative \(\frac{dx}{dt}\) represents the rate of horizontal movement, while \(\frac{dy}{dt}\) represents the rate of vertical movement. Since these components are derived from trigonometric functions involving \(\theta\), the chain rule is employed, coupling the derivative of \(\theta\) with respect to time (\(\frac{d\theta}{dt}\)) into the expression. This yields the velocity components in the horizontal and vertical directions, which are crucial for understanding the dynamics of motion in a circular path.

Theta as a Function of Time

The angle \(\theta\) as a function of time is the mathematical representation of how the angle changes as time progresses. For uniform circular motion, \(\theta\) increases linearly with time since the rate of rotation is constant. The general relationship is \(\theta(t) = \omega t + \theta_0\), where \(\omega\) is the angular velocity (constant in uniform motion), and \(\theta_0\) is the initial angle at \(t = 0\).

In the presented exercise, the wheel makes 8 revolutions per second. Since each revolution corresponds to an angle of \(2\pi\) radians, the function \(\theta(t)\) for this wheel is \(\theta(t) = 16\pi t\), explicitly stating that the angle in radians is 16\pi times the elapsed time in seconds, starting from an initial angle of zero.

Horizontal and Vertical Movement Rate

The rates of horizontal and vertical movement refer to the speeds at which a point is moving along the x-axis and y-axis, respectively. Using differentiation, we can find these rates for any position on the circle characterized by an angle \(\theta\). As exemplified in the exercise, the derivative \(\frac{dx}{dt}\) at a given \(\theta\) measures the horizontal velocity, while \(\frac{dy}{dt}\) measures the vertical velocity.

For specific angles like \(\theta =\pi/4\), \(\pi/2\), and \(\pi\), we substitute the angle into the differentiated equations to find the horizontal and vertical speeds. For instance, at \(\theta = \pi/4\), a point on the wheel is moving horizontally at \(\frac{dx}{dt}\) and vertically at \(\frac{dy}{dt}\), which are evaluated using \(\sin(\pi/4) = \frac{\sqrt{2}}{2}\) and \(\cos(\pi/4) = \frac{\sqrt{2}}{2}\), respectively. These calculations allow us to analyze the motion of the wheel at various positions.